Question

How do you solve `abs(5 / (2x-1))>= abs(1 / (x - 2))`?

Answer 1

`x<1/2` or `1/2< x <=11/7` or `x >=3`

Explanation:

At first observe that `xne 2` and `xne 1/2` by crossmultiplication we get

`5|x-2|>=|2x-1|` no we distinguish several cases:

a) `x>2` then we get

`5x-10>=2x-1` and we get `x>=3`

b)`1/2<x<2` and we get

`10-5x>=2x-1` and we get `x<11/7`

`1/2<x<=11/7`

in our last case we have

c)`x<1/2`

and we get

`10-5x>=1-2x`

so `x<1/2` and we get the interval like above.

Answer 2

The solution is `x in (-oo,1/2) uu(1/2,11/7]uu[3,+oo)`

Explanation:

You cannot cross multiply when you have an inequality

The inequality is

`|5/(2x-1)|>=|1/(x-2)|`

`<=>`, `5/|2x-1|>=1/|(x-2)|`

`<=>`, `5/|2x-1|-1/|(x-2)|>=0`

There are `2` points to consider

`2x-1=0`, `=>`, `x=1/2` and

`x-2=0`, `=>`, `x=2`

There are `3` intervals to consider

`I_1=(-oo, 1/2)` and `I_2=(1/2, 2)` and `I_3=(2,+oo)`

In the first interval `I_1`, we have

`5/(-2x+1)-1/(-x+2)>=0`

`(5(2-x)-1(1-2x))/((1-2x)(2-x))>=0`

`(10-5x-1+2x)/((1-2x)(2-x))>=0`

`(9-3x)/((1-2x)(2-x))>=0`

`(3(3-x))/((1-2x)(2-x))>=0`

Let `f(x)=(3(3-x))/((1-2x)(2-x))`

Solving this equation with a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `1/2` `color(white)(aaaaaa)` `2` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `1-2x` `color(white)(aaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `-` `color(white)(aaa)` `-`

`color(white)(aaaa)` `2-x` `color(white)(aaaa)` `+` `color(white)(aaaa)` ###color(white)(aaaa)#`+` `color(white)(aa)` `||` `color(white)(a)` `-` `color(white)(aaa)` `-`

`color(white)(aaaa)` `3-x` `color(white)(aaaa)` `+` `color(white)(aaaa)` ###color(white)(aaaa)#`+` `color(white)(aaaa)` `+` `color(white)(a)` `0` `color(white)(a)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-` `color(white)(a)` `||` `color(white)(a)` `+` `color(white)(a)` `0` `color(white)(a)` `-`

Therefore,

In the interval `I_1`, `f(x)>=0`,when `x in (-oo,1/2)`

In the second interval `I_2=(1/2,2)`

`5/(2x-1)-1/(-x+2)>=0`

`<=>`, `(5(2-x)-1(2x-1))/((2x-1)(2-x))>=0`

`<=>`, `(10-5x-2x+1)/((2x-1)(2-x))>=0`

`<=>`, `(11-7x)/((2x-1)(2-x))>=0`

Let `g(x)=(11-7x)/((2x-1)(2-x))`

Solving this inequality with a sign chart,

`g(x)>=0` when `x in (1/2,11/7]`

In the third interval `I_2=(2, +oo)`

`5/(2x-1)-1/(x-2)>=0`

`<=>`, `(5(x-2)-1(2x-1))/((2x-1)(x-2))>=0`

`<=>`, `((5x-10-2x+1))/((2x-1)(x-2))>=0`

`<=>`, `((3x-9))/((2x-1)(x-2))>=0`

`<=>`, `(3(x-3))/((2x-1)(x-2))>=0`

`h(x)=(3(x-3))/((2x-1)(x-2))`

Solving this inequality with a sign chart,

`h(x)>=0` when `x in [3,+oo)`

graph{5/(|2x-1|)-1/(|x-2|) [-14.24, 14.24, -7.12, 7.12]}